Question
Use Kirchhoff ’s rules to determine the potential difference between the points A and D when no current flows in the arm BE of the electric network shown in the figure.
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Answer
According to Kirchoff‟s Junction law at B
$\text{i}_{3} = \text{i}_{1} + \text{i}_{2} \therefore\text{i}_{3} = \text{i}_{1}$
(As I2=0 (given)
Applying second law to loop AFEB
$\text{i}_{3}\times2 + \text{i}_{3} \times3 + \text{i}_{2}\text{R}_{1} = 1+ 3 + 6 $
$\therefore\text{i}_{3} = \text{i}_{1} = 2 \text{A}$
From A to D along AFD $\therefore\text{V}_{\text{AD}} = 2\text{i}_{3} -1 +3 \times\text{i}_{3}$
$ = ( 4- 1 + 6 ) \text{V}$
= 9 V.
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